Explain Poisson's ratio and show that its value is less than $0.5$.

(N/A) Poisson's ratio is defined as the ratio of lateral strain to longitudinal strain.
Let a cylindrical rod have length $l$ and radius $r$. When the rod is stretched,its length increases by $\Delta l$ and its radius decreases by $\Delta r$.
The lateral strain is $\epsilon_{lat} = -\frac{\Delta r}{r}$ and the longitudinal strain is $\epsilon_{long} = \frac{\Delta l}{l}$.
Poisson's ratio $\mu$ is given by $\mu = \frac{\epsilon_{lat}}{\epsilon_{long}} = -\frac{\Delta r / r}{\Delta l / l}$.
Thus,$\frac{\Delta r}{r} = -\mu \frac{\Delta l}{l} \quad \dots (1)$
The volume $V$ of the rod is $V = \pi r^2 l$.
Differentiating to find the change in volume:
$\frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac{\Delta l}{l} \quad \dots (2)$
Substituting equation $(1)$ into $(2)$:
$\frac{\Delta V}{V} = 2(-\mu \frac{\Delta l}{l}) + \frac{\Delta l}{l} = \frac{\Delta l}{l} (1 - 2\mu)$
Since the volume of a material cannot increase when it is stretched $(\Delta V \ge 0)$,we must have $(1 - 2\mu) \ge 0$.
Therefore,$1 \ge 2\mu$,which implies $\mu \le 0.5$.